Finding the derivative of an exponential function without the chain rule Given $f(x) = a^x$, one can compute the derivative of $f(x)$ using the chain rule quickly by noticing that
$f(x) = a^x = \left(e^{ln(a)}\right)^x$.
But how would you go about computing the derivative of $f(x)$ without the chain rule, with just the $f'(x) = \lim_\limits{h \to 0} \frac{f(x + h) - f(x)}{h}$ or the $f'(x) = \lim_\limits{x \to a} \frac{f(x) - f(a)}{x - a}$ definition of a derivative and the fact that $\lim_\limits{h \to 0} \frac{e^h - 1}{h} = 1$?
$f'(x) = \lim_\limits{h \to 0} \frac{a^{x + h} - a^x}{h} = a^x \cdot \lim_\limits{h \to 0} \frac{a^h - 1}{h} = a^x \cdot \lim_\limits{h \to 0} \frac{\left(e^{ln(a)}\right)^h - 1}{h} = a^x \cdot \lim_\limits{h \to 0} \frac{(e^h)^{ln(a)} - 1}{h}$.
How would you proceed from here?
 A: Here is a way to show $a^x$ has derivative $\ln(a) \cdot a^x$ without using chain rule. It uses the series expansion of $e^x$.
Since $a^x = e^{x \ln(a)}$, we have:
$$
a^x = \sum_{n=0}^\infty \frac{(x\ln(a))^n}{n!}.
$$
We can take the derivative term by term (using your favorite convergence theorem), which yields:
$$
\frac{d}{dx} a^x = \sum_{n=0}^\infty \frac{nx^{n-1} \ln(a)^n}{n!}.
$$
Now we make the very important observation that the first term (when $n=0$) is zero. Hence, we can index starting from $n = 1$ and cancel out the $n$ in the numerator:
$$
\frac{d}{dx} a^x = \sum_{n=1}^\infty \frac{x^{n-1}\ln(a)^n}{(n-1)!} = \ln(a) \cdot \sum^\infty_{n=1} \frac{x^{n-1}\ln(a)^{n-1}}{(n-1)!},
$$
noting we can pull out the factor of $\ln(a)$ from every term now. But the last sum can now be re-indexed,
$$
\sum^\infty_{n=1} \frac{x^{n-1}\ln(a)^{n-1}}{(n-1)!} = \sum_{n=0}^\infty \frac{x^n\ln(a)^n}{n!} = a^x,
$$
so in conclusion we have:
$$
\frac{d}{dx} a^x = \ln(a) \cdot a^x.
$$
A: Here is a way that uses our knowledge that $\lim_{h\to0}\frac{\mathrm e^h-1}{h}=1$.
At some point in the calculation you demonstrate, we get to the part where we need to evaluate
$$\lim_{h\to0}\frac{\left(\mathrm e^{\ln(a)}\right)^h-1}{h}.$$
Here we can use $(\mathrm{e}^{\ln a})^h=\mathrm e^{h\ln a}$ and substitute $u:=h\ln a$ and note that $u\to0\Leftrightarrow h\to0$. So we are left evaluating
$$\begin{align}\lim_{u\to0}\frac{\mathrm e^u-1}{\frac{u}{\ln a}}&=\lim_{u\to0}\ln(a)\frac{\mathrm e^u-1}{u}\\
&=\ln a\lim_{u\to0}\frac{\mathrm e^u-1}{u}\\
&=\ln (a)\cdot 1\\
&=\ln (a).
\end{align}$$
A: You have reduced the problem to showing $$
\ln a=\lim_{h\rightarrow 0}\frac{(e^h)^{\ln a}-1}{h}
$$
To do this, note $(e^h)^{\ln a}=e^{h\ln a}$, and make the substitution $h= h'/\ln a$. Then we obtain $$\lim_{h\rightarrow 0}\frac{(e^h)^{\ln a}-1}{h}=\lim_{h'\rightarrow 0}\frac{e^{(h'/\ln a)\ln a}-1}{h'/\ln a}=\lim_{h'\rightarrow 0}\frac{e^{h'}-1}{h'}\ln a=\ln a
$$
A: So it boils down to finding this limit:
$$\lim_\limits{h \to 0} \frac{a^{h} - 1}{h}$$
Let $y=\frac{a^{h} - 1}{h}$, then $h=\frac{\ln(1+y)}{\ln a}$. Note that as $h\rightarrow0$, so does $y.$  Now substitute this value to get
$$\lim_\limits{h \to 0} \frac{a^{h} - 1}{h}=\ln a \lim_\limits{y \to 0}  \frac{y}{\ln(1+y)}=\ln a \lim_\limits{y \to 0}  \frac{1}{\frac{1}{y}\ln(1+y)}==\ln a \lim_\limits{y \to 0}  \frac{1}{{\ln(1+y)}^\frac{1}{y}}=\frac{\ln a}{\ln e}=\ln(a).$$
Note that at the end, we used the fact that $\ln$ is continuous for positive inputs and that $\lim_{y\rightarrow 0}{(1+y)}^\frac{1}{y}=e.$
Finally
$$f'(x)=a^x\lim_\limits{h \to 0} \frac{a^{h} - 1}{h}=\ln(a)\cdot a^x$$
